Then, if \(|a|<1\)  the solution \( \mathring{H}^{\lbrace 1\rbrace }(x)\)  converges inside
the circle \(|x|=1\)  which contains \(x=0\)  and \(x=a\) ; by the Raabe's test
[1]}, [2]}
it converges also on \(|x|=1\)  if \(\mbox{Re}\;\delta <1\) .
On the other hand, if \(|a|>1\)  the solution
converges inside
the circle \(|x|=|a|\)  which contains \(x=0\)  and \(x=1\) ; by the Raabe's test
it converges also on \(|x|=|a|\)  if \(\mbox{Re}\;\epsilon <1\) .
In sum,
\(\begin{array}{c}\mbox{for minimal}\\\mbox{solution}\end{array}{\left\lbrace \begin{array}{ll}\mbox{if } |a|<1,\; \mathring{H}^{(1)}(x) \mbox{ convergesfor } |x|<1 \mbox{ and on } |x|=1 \mbox{ if }\mbox{Re}\;\delta <1; \vspace{5.69054pt}\\\mbox{if } |a|>1, \; \mathring{H}^{(1)}(x) \mbox{ convergesfor } |x|<|a| \mbox{ and on } |x|=|a| \mbox{ if }\mbox{Re}\;\epsilon <1.\end{array}\right.}\) 
